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not parallel; further let VP and YQ be the velocities. Dn the line PO perpendicular to VP in a plane parallel to t guide plane; also draw QO perpendicular to VQ in the sai plane. Then the instantaneous axis passes through 0, t point of intersection, and is perpendicular to the plane PO
1. Discuss the motion of a uniform circular cylinder which rolls do a rough inclined plane without slipping.
FIG. 115.
Let m = the mass of the cylinder.
I = the moment of inertia of the cylinder about the element
a = the radius of the cylinder. v  the velocity of the axis of the cylinder, co = the angular velocity of the cylinder.
TORQUE METHOD.  The torque is due to the weight of the cylinder and the reaction of the plane. It equals the moment of the weight about the element of contact. Therefore
G = mga sin a..
Substituting this value of G in the torque equation we have Jcb = mga sin a.
ENERGY METHOD.  In movi through a distance s along 1 plane the potential energy of 1 cylinder is diminished by an amou equal to
mgh = mgs sin a. Therefore the energy equation gh
17^2 __ i ju0z = mgs sin a. Differentiating with respect to t time
Jcoo? = mgs sin a. Jco = mga sin a.